Consider a representation of \(\Vect{0}\) as a linear combination of pairwise distinct vectors in an orthogonal set \(S\):

\(t_1 \Vect{a}_1 + \cdots + t_m \Vect{a}_m\)

\(=\)

\(\Vect{0}\)

We need to show that \(t_1=\cdots = t_m =0\). To test the value of \(t_i\), we take the dot product of both sides of this equation by the vector \(\Vect{a}_i\), and find

\(t_1(\DotPr{ \Vect{a}_1 }{ \Vect{a}_i }) + \cdots + t_i(\DotPr{ \Vect{a}_i }{ \Vect{a}_i })+ \cdots + t_m(\DotPr{ \Vect{a}_m }{ \Vect{a}_i })\)	\(=\)	\(\DotPr{ \Vect{0} }{ \Vect{a}_i }\)
\(t_i(\DotPr{ \Vect{a}_i }{ \Vect{a}_i })\)	\(= \)	\(0\)

We have \(\DotPr{ \Vect{a}_i }{ \Vect{a}_i } \neq 0\) because \(\Vect{a}_i\neq \Vect{0}\) and, therefore, \(t_i=0\). This implies that \(S\) is linearly independent.